Question

Find the derivatives of sin2nx÷cos^2nx​

Answer 1

Topic :-

Differentiation

To Differentiate :-

$y=\dfrac{\sin (2nx)}{\cos^2(nx)}$

Solution :-

$y=\dfrac{\sin (2nx)}{\cos^2(nx)}$

$y=\dfrac{2\sin (nx)\cdot\cos (nx)}{\cos^2(nx)}$

$(\because \sin2\theta=2\sin\theta\cdot\cos\theta)$

$y=\dfrac{2\sin (nx)}{\cos(nx)}$

$y=2\tan (nx)$

$\left( \because \dfrac{\sin \theta}{\cos \theta}=\tan \theta\right)$

$\dfrac{dy}{dx}=\dfrac{d(2\tan (nx))}{dx}$

$\dfrac{dy}{dx}=2\cdot\dfrac{d(\tan (nx))}{dx}$

$\left( \because \dfrac{d(k\cdot f(x))}{dx}=k\cdot\dfrac{d(f(x))}{dx},where\:k\:is\:a\:constant.\right)$

$\dfrac{dy}{dx}=2\cdot \sec^2(nx)\cdot\dfrac{d(nx)}{dx}$

$\left(\because \dfrac{d(\tan t)}{dx}=\sec^2t\cdot \dfrac{dt}{dx}\right)$

$\dfrac{dy}{dx}=2\cdot \sec^2(nx)\cdot n\cdot \dfrac{dx}{dx}$

$\left( \because \dfrac{d(k\cdot f(x))}{dx}=k\cdot\dfrac{d(f(x))}{dx},where\:k\:is\:a\:constant.\right)$

$\dfrac{dy}{dx}=2n\cdot \sec^2(nx)$

$\left(\because \dfrac{dx}{dx}=1 \right)$

Answer :-

$\underline{\boxed{\dfrac{dy}{dx}=2n\cdot \sec^2(nx)}}$